Left Termination proof of /tmp/tmpT37f70/permutation.pl

Left Termination of the query pattern perm_in_2(g, a) w.r.t. the given Prolog program could successfully be proven:

↳ Prolog

  ↳ PrologToPiTRSProof

Clauses:

app1(.(X0, X), Y, .(X0, Z)) :- app1(X, Y, Z).
app1([], Y, Y).
app2(.(X0, X), Y, .(X0, Z)) :- app2(X, Y, Z).
app2([], Y, Y).
perm(X, .(X0, Y)) :- ','(app1(X1, .(X0, X2), X), ','(app2(X1, X2, Z), perm(Z, Y))).
perm([], []).

Queries:

perm(g,a).

We use the technique of [30].Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

perm_in([], []) → perm_out([], [])
perm_in(X, .(X0, Y)) → U3(X, X0, Y, app1_in(X1, .(X0, X2), X))
app1_in([], Y, Y) → app1_out([], Y, Y)
app1_in(.(X0, X), Y, .(X0, Z)) → U1(X0, X, Y, Z, app1_in(X, Y, Z))
U1(X0, X, Y, Z, app1_out(X, Y, Z)) → app1_out(.(X0, X), Y, .(X0, Z))
U3(X, X0, Y, app1_out(X1, .(X0, X2), X)) → U4(X, X0, Y, X1, X2, app2_in(X1, X2, Z))
app2_in([], Y, Y) → app2_out([], Y, Y)
app2_in(.(X0, X), Y, .(X0, Z)) → U2(X0, X, Y, Z, app2_in(X, Y, Z))
U2(X0, X, Y, Z, app2_out(X, Y, Z)) → app2_out(.(X0, X), Y, .(X0, Z))
U4(X, X0, Y, X1, X2, app2_out(X1, X2, Z)) → U5(X, X0, Y, perm_in(Z, Y))
U5(X, X0, Y, perm_out(Z, Y)) → perm_out(X, .(X0, Y))

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

Pi-finite rewrite system:
The TRS R consists of the following rules:

perm_in([], []) → perm_out([], [])
perm_in(X, .(X0, Y)) → U3(X, X0, Y, app1_in(X1, .(X0, X2), X))
app1_in([], Y, Y) → app1_out([], Y, Y)
app1_in(.(X0, X), Y, .(X0, Z)) → U1(X0, X, Y, Z, app1_in(X, Y, Z))
U1(X0, X, Y, Z, app1_out(X, Y, Z)) → app1_out(.(X0, X), Y, .(X0, Z))
U3(X, X0, Y, app1_out(X1, .(X0, X2), X)) → U4(X, X0, Y, X1, X2, app2_in(X1, X2, Z))
app2_in([], Y, Y) → app2_out([], Y, Y)
app2_in(.(X0, X), Y, .(X0, Z)) → U2(X0, X, Y, Z, app2_in(X, Y, Z))
U2(X0, X, Y, Z, app2_out(X, Y, Z)) → app2_out(.(X0, X), Y, .(X0, Z))
U4(X, X0, Y, X1, X2, app2_out(X1, X2, Z)) → U5(X, X0, Y, perm_in(Z, Y))
U5(X, X0, Y, perm_out(Z, Y)) → perm_out(X, .(X0, Y))

PERM_IN(X, .(X0, Y)) → U3¹(X, X0, Y, app1_in(X1, .(X0, X2), X))
PERM_IN(X, .(X0, Y)) → APP1_IN(X1, .(X0, X2), X)
APP1_IN(.(X0, X), Y, .(X0, Z)) → U1¹(X0, X, Y, Z, app1_in(X, Y, Z))
APP1_IN(.(X0, X), Y, .(X0, Z)) → APP1_IN(X, Y, Z)
U3¹(X, X0, Y, app1_out(X1, .(X0, X2), X)) → U4¹(X, X0, Y, X1, X2, app2_in(X1, X2, Z))
U3¹(X, X0, Y, app1_out(X1, .(X0, X2), X)) → APP2_IN(X1, X2, Z)
APP2_IN(.(X0, X), Y, .(X0, Z)) → U2¹(X0, X, Y, Z, app2_in(X, Y, Z))
APP2_IN(.(X0, X), Y, .(X0, Z)) → APP2_IN(X, Y, Z)
U4¹(X, X0, Y, X1, X2, app2_out(X1, X2, Z)) → U5¹(X, X0, Y, perm_in(Z, Y))
U4¹(X, X0, Y, X1, X2, app2_out(X1, X2, Z)) → PERM_IN(Z, Y)

The TRS R consists of the following rules:

perm_in([], []) → perm_out([], [])
perm_in(X, .(X0, Y)) → U3(X, X0, Y, app1_in(X1, .(X0, X2), X))
app1_in([], Y, Y) → app1_out([], Y, Y)
app1_in(.(X0, X), Y, .(X0, Z)) → U1(X0, X, Y, Z, app1_in(X, Y, Z))
U1(X0, X, Y, Z, app1_out(X, Y, Z)) → app1_out(.(X0, X), Y, .(X0, Z))
U3(X, X0, Y, app1_out(X1, .(X0, X2), X)) → U4(X, X0, Y, X1, X2, app2_in(X1, X2, Z))
app2_in([], Y, Y) → app2_out([], Y, Y)
app2_in(.(X0, X), Y, .(X0, Z)) → U2(X0, X, Y, Z, app2_in(X, Y, Z))
U2(X0, X, Y, Z, app2_out(X, Y, Z)) → app2_out(.(X0, X), Y, .(X0, Z))
U4(X, X0, Y, X1, X2, app2_out(X1, X2, Z)) → U5(X, X0, Y, perm_in(Z, Y))
U5(X, X0, Y, perm_out(Z, Y)) → perm_out(X, .(X0, Y))

The argument filtering Pi contains the following mapping:
perm_in(x1, x2) = perm_in(x1)
[] = []
perm_out(x1, x2) = perm_out(x2)
.(x1, x2) = .(x2)
U3(x1, x2, x3, x4) = U3(x4)
app1_in(x1, x2, x3) = app1_in(x3)
app1_out(x1, x2, x3) = app1_out(x1, x2)
U1(x1, x2, x3, x4, x5) = U1(x5)
U4(x1, x2, x3, x4, x5, x6) = U4(x6)
app2_in(x1, x2, x3) = app2_in(x1, x2)
app2_out(x1, x2, x3) = app2_out(x3)
U2(x1, x2, x3, x4, x5) = U2(x5)
U5(x1, x2, x3, x4) = U5(x4)
U5¹(x1, x2, x3, x4) = U5¹(x4)
U3¹(x1, x2, x3, x4) = U3¹(x4)
APP2_IN(x1, x2, x3) = APP2_IN(x1, x2)
U4¹(x1, x2, x3, x4, x5, x6) = U4¹(x6)
PERM_IN(x1, x2) = PERM_IN(x1)
U2¹(x1, x2, x3, x4, x5) = U2¹(x5)
U1¹(x1, x2, x3, x4, x5) = U1¹(x5)
APP1_IN(x1, x2, x3) = APP1_IN(x3)

We have to consider all (P,R,Pi)-chains

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

Pi DP problem:
The TRS P consists of the following rules:

PERM_IN(X, .(X0, Y)) → U3¹(X, X0, Y, app1_in(X1, .(X0, X2), X))
PERM_IN(X, .(X0, Y)) → APP1_IN(X1, .(X0, X2), X)
APP1_IN(.(X0, X), Y, .(X0, Z)) → U1¹(X0, X, Y, Z, app1_in(X, Y, Z))
APP1_IN(.(X0, X), Y, .(X0, Z)) → APP1_IN(X, Y, Z)
U3¹(X, X0, Y, app1_out(X1, .(X0, X2), X)) → U4¹(X, X0, Y, X1, X2, app2_in(X1, X2, Z))
U3¹(X, X0, Y, app1_out(X1, .(X0, X2), X)) → APP2_IN(X1, X2, Z)
APP2_IN(.(X0, X), Y, .(X0, Z)) → U2¹(X0, X, Y, Z, app2_in(X, Y, Z))
APP2_IN(.(X0, X), Y, .(X0, Z)) → APP2_IN(X, Y, Z)
U4¹(X, X0, Y, X1, X2, app2_out(X1, X2, Z)) → U5¹(X, X0, Y, perm_in(Z, Y))
U4¹(X, X0, Y, X1, X2, app2_out(X1, X2, Z)) → PERM_IN(Z, Y)

The TRS R consists of the following rules:

perm_in([], []) → perm_out([], [])
perm_in(X, .(X0, Y)) → U3(X, X0, Y, app1_in(X1, .(X0, X2), X))
app1_in([], Y, Y) → app1_out([], Y, Y)
app1_in(.(X0, X), Y, .(X0, Z)) → U1(X0, X, Y, Z, app1_in(X, Y, Z))
U1(X0, X, Y, Z, app1_out(X, Y, Z)) → app1_out(.(X0, X), Y, .(X0, Z))
U3(X, X0, Y, app1_out(X1, .(X0, X2), X)) → U4(X, X0, Y, X1, X2, app2_in(X1, X2, Z))
app2_in([], Y, Y) → app2_out([], Y, Y)
app2_in(.(X0, X), Y, .(X0, Z)) → U2(X0, X, Y, Z, app2_in(X, Y, Z))
U2(X0, X, Y, Z, app2_out(X, Y, Z)) → app2_out(.(X0, X), Y, .(X0, Z))
U4(X, X0, Y, X1, X2, app2_out(X1, X2, Z)) → U5(X, X0, Y, perm_in(Z, Y))
U5(X, X0, Y, perm_out(Z, Y)) → perm_out(X, .(X0, Y))

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

              ↳ PiDP

Pi DP problem:
The TRS P consists of the following rules:

APP2_IN(.(X0, X), Y, .(X0, Z)) → APP2_IN(X, Y, Z)

The TRS R consists of the following rules:

perm_in([], []) → perm_out([], [])
perm_in(X, .(X0, Y)) → U3(X, X0, Y, app1_in(X1, .(X0, X2), X))
app1_in([], Y, Y) → app1_out([], Y, Y)
app1_in(.(X0, X), Y, .(X0, Z)) → U1(X0, X, Y, Z, app1_in(X, Y, Z))
U1(X0, X, Y, Z, app1_out(X, Y, Z)) → app1_out(.(X0, X), Y, .(X0, Z))
U3(X, X0, Y, app1_out(X1, .(X0, X2), X)) → U4(X, X0, Y, X1, X2, app2_in(X1, X2, Z))
app2_in([], Y, Y) → app2_out([], Y, Y)
app2_in(.(X0, X), Y, .(X0, Z)) → U2(X0, X, Y, Z, app2_in(X, Y, Z))
U2(X0, X, Y, Z, app2_out(X, Y, Z)) → app2_out(.(X0, X), Y, .(X0, Z))
U4(X, X0, Y, X1, X2, app2_out(X1, X2, Z)) → U5(X, X0, Y, perm_in(Z, Y))
U5(X, X0, Y, perm_out(Z, Y)) → perm_out(X, .(X0, Y))

The argument filtering Pi contains the following mapping:
perm_in(x1, x2) = perm_in(x1)
[] = []
perm_out(x1, x2) = perm_out(x2)
.(x1, x2) = .(x2)
U3(x1, x2, x3, x4) = U3(x4)
app1_in(x1, x2, x3) = app1_in(x3)
app1_out(x1, x2, x3) = app1_out(x1, x2)
U1(x1, x2, x3, x4, x5) = U1(x5)
U4(x1, x2, x3, x4, x5, x6) = U4(x6)
app2_in(x1, x2, x3) = app2_in(x1, x2)
app2_out(x1, x2, x3) = app2_out(x3)
U2(x1, x2, x3, x4, x5) = U2(x5)
U5(x1, x2, x3, x4) = U5(x4)
APP2_IN(x1, x2, x3) = APP2_IN(x1, x2)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting [30] we can delete all non-usable rules from R.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

              ↳ PiDP

Pi DP problem:
The TRS P consists of the following rules:

APP2_IN(.(X0, X), Y, .(X0, Z)) → APP2_IN(X, Y, Z)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x2)
APP2_IN(x1, x2, x3) = APP2_IN(x1, x2)

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem [30] into ordinary QDP problem [15] by application of Pi.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ PiDP

              ↳ PiDP

Q DP problem:
The TRS P consists of the following rules:

APP2_IN(.(X), Y) → APP2_IN(X, Y)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

APP2_IN(.(X), Y) → APP2_IN(X, Y)
The graph contains the following edges 1 > 1, 2 >= 2

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

Pi DP problem:
The TRS P consists of the following rules:

APP1_IN(.(X0, X), Y, .(X0, Z)) → APP1_IN(X, Y, Z)

The TRS R consists of the following rules:

perm_in([], []) → perm_out([], [])
perm_in(X, .(X0, Y)) → U3(X, X0, Y, app1_in(X1, .(X0, X2), X))
app1_in([], Y, Y) → app1_out([], Y, Y)
app1_in(.(X0, X), Y, .(X0, Z)) → U1(X0, X, Y, Z, app1_in(X, Y, Z))
U1(X0, X, Y, Z, app1_out(X, Y, Z)) → app1_out(.(X0, X), Y, .(X0, Z))
U3(X, X0, Y, app1_out(X1, .(X0, X2), X)) → U4(X, X0, Y, X1, X2, app2_in(X1, X2, Z))
app2_in([], Y, Y) → app2_out([], Y, Y)
app2_in(.(X0, X), Y, .(X0, Z)) → U2(X0, X, Y, Z, app2_in(X, Y, Z))
U2(X0, X, Y, Z, app2_out(X, Y, Z)) → app2_out(.(X0, X), Y, .(X0, Z))
U4(X, X0, Y, X1, X2, app2_out(X1, X2, Z)) → U5(X, X0, Y, perm_in(Z, Y))
U5(X, X0, Y, perm_out(Z, Y)) → perm_out(X, .(X0, Y))

The argument filtering Pi contains the following mapping:
perm_in(x1, x2) = perm_in(x1)
[] = []
perm_out(x1, x2) = perm_out(x2)
.(x1, x2) = .(x2)
U3(x1, x2, x3, x4) = U3(x4)
app1_in(x1, x2, x3) = app1_in(x3)
app1_out(x1, x2, x3) = app1_out(x1, x2)
U1(x1, x2, x3, x4, x5) = U1(x5)
U4(x1, x2, x3, x4, x5, x6) = U4(x6)
app2_in(x1, x2, x3) = app2_in(x1, x2)
app2_out(x1, x2, x3) = app2_out(x3)
U2(x1, x2, x3, x4, x5) = U2(x5)
U5(x1, x2, x3, x4) = U5(x4)
APP1_IN(x1, x2, x3) = APP1_IN(x3)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting [30] we can delete all non-usable rules from R.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

Pi DP problem:
The TRS P consists of the following rules:

APP1_IN(.(X0, X), Y, .(X0, Z)) → APP1_IN(X, Y, Z)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x2)
APP1_IN(x1, x2, x3) = APP1_IN(x3)

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem [30] into ordinary QDP problem [15] by application of Pi.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ PiDP

Q DP problem:
The TRS P consists of the following rules:

APP1_IN(.(Z)) → APP1_IN(Z)

From the DPs we obtained the following set of size-change graphs:

APP1_IN(.(Z)) → APP1_IN(Z)
The graph contains the following edges 1 > 1

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

Pi DP problem:
The TRS P consists of the following rules:

U4¹(X, X0, Y, X1, X2, app2_out(X1, X2, Z)) → PERM_IN(Z, Y)
U3¹(X, X0, Y, app1_out(X1, .(X0, X2), X)) → U4¹(X, X0, Y, X1, X2, app2_in(X1, X2, Z))
PERM_IN(X, .(X0, Y)) → U3¹(X, X0, Y, app1_in(X1, .(X0, X2), X))

The TRS R consists of the following rules:

perm_in([], []) → perm_out([], [])
perm_in(X, .(X0, Y)) → U3(X, X0, Y, app1_in(X1, .(X0, X2), X))
app1_in([], Y, Y) → app1_out([], Y, Y)
app1_in(.(X0, X), Y, .(X0, Z)) → U1(X0, X, Y, Z, app1_in(X, Y, Z))
U1(X0, X, Y, Z, app1_out(X, Y, Z)) → app1_out(.(X0, X), Y, .(X0, Z))
U3(X, X0, Y, app1_out(X1, .(X0, X2), X)) → U4(X, X0, Y, X1, X2, app2_in(X1, X2, Z))
app2_in([], Y, Y) → app2_out([], Y, Y)
app2_in(.(X0, X), Y, .(X0, Z)) → U2(X0, X, Y, Z, app2_in(X, Y, Z))
U2(X0, X, Y, Z, app2_out(X, Y, Z)) → app2_out(.(X0, X), Y, .(X0, Z))
U4(X, X0, Y, X1, X2, app2_out(X1, X2, Z)) → U5(X, X0, Y, perm_in(Z, Y))
U5(X, X0, Y, perm_out(Z, Y)) → perm_out(X, .(X0, Y))

The argument filtering Pi contains the following mapping:
perm_in(x1, x2) = perm_in(x1)
[] = []
perm_out(x1, x2) = perm_out(x2)
.(x1, x2) = .(x2)
U3(x1, x2, x3, x4) = U3(x4)
app1_in(x1, x2, x3) = app1_in(x3)
app1_out(x1, x2, x3) = app1_out(x1, x2)
U1(x1, x2, x3, x4, x5) = U1(x5)
U4(x1, x2, x3, x4, x5, x6) = U4(x6)
app2_in(x1, x2, x3) = app2_in(x1, x2)
app2_out(x1, x2, x3) = app2_out(x3)
U2(x1, x2, x3, x4, x5) = U2(x5)
U5(x1, x2, x3, x4) = U5(x4)
U3¹(x1, x2, x3, x4) = U3¹(x4)
U4¹(x1, x2, x3, x4, x5, x6) = U4¹(x6)
PERM_IN(x1, x2) = PERM_IN(x1)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting [30] we can delete all non-usable rules from R.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

Pi DP problem:
The TRS P consists of the following rules:

U4¹(X, X0, Y, X1, X2, app2_out(X1, X2, Z)) → PERM_IN(Z, Y)
U3¹(X, X0, Y, app1_out(X1, .(X0, X2), X)) → U4¹(X, X0, Y, X1, X2, app2_in(X1, X2, Z))
PERM_IN(X, .(X0, Y)) → U3¹(X, X0, Y, app1_in(X1, .(X0, X2), X))

The TRS R consists of the following rules:

app2_in([], Y, Y) → app2_out([], Y, Y)
app2_in(.(X0, X), Y, .(X0, Z)) → U2(X0, X, Y, Z, app2_in(X, Y, Z))
app1_in([], Y, Y) → app1_out([], Y, Y)
app1_in(.(X0, X), Y, .(X0, Z)) → U1(X0, X, Y, Z, app1_in(X, Y, Z))
U2(X0, X, Y, Z, app2_out(X, Y, Z)) → app2_out(.(X0, X), Y, .(X0, Z))
U1(X0, X, Y, Z, app1_out(X, Y, Z)) → app1_out(.(X0, X), Y, .(X0, Z))

The argument filtering Pi contains the following mapping:
[] = []
.(x1, x2) = .(x2)
app1_in(x1, x2, x3) = app1_in(x3)
app1_out(x1, x2, x3) = app1_out(x1, x2)
U1(x1, x2, x3, x4, x5) = U1(x5)
app2_in(x1, x2, x3) = app2_in(x1, x2)
app2_out(x1, x2, x3) = app2_out(x3)
U2(x1, x2, x3, x4, x5) = U2(x5)
U3¹(x1, x2, x3, x4) = U3¹(x4)
U4¹(x1, x2, x3, x4, x5, x6) = U4¹(x6)
PERM_IN(x1, x2) = PERM_IN(x1)

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem [30] into ordinary QDP problem [15] by application of Pi.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ RuleRemovalProof

Q DP problem:
The TRS P consists of the following rules:

U3¹(app1_out(X1, .(X2))) → U4¹(app2_in(X1, X2))
PERM_IN(X) → U3¹(app1_in(X))
U4¹(app2_out(Z)) → PERM_IN(Z)

The TRS R consists of the following rules:

app2_in([], Y) → app2_out(Y)
app2_in(.(X), Y) → U2(app2_in(X, Y))
app1_in(Y) → app1_out([], Y)
app1_in(.(Z)) → U1(app1_in(Z))
U2(app2_out(Z)) → app2_out(.(Z))
U1(app1_out(X, Y)) → app1_out(.(X), Y)

The set Q consists of the following terms:

app2_in(x0, x1)
app1_in(x0)
U2(x0)
U1(x0)

We have to consider all (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

U3¹(app1_out(X1, .(X2))) → U4¹(app2_in(X1, X2))

Strictly oriented rules of the TRS R:

app1_in(Y) → app1_out([], Y)

Used ordering: POLO with Polynomial interpretation [25]:

POL(.(x₁)) = 1 + x₁
POL(PERM_IN(x₁)) = 1 + 2·x₁
POL(U1(x₁)) = 2 + x₁
POL(U2(x₁)) = 2 + x₁
POL(U3¹(x₁)) = x₁
POL(U4¹(x₁)) = 1 + x₁
POL([]) = 0
POL(app1_in(x₁)) = 1 + 2·x₁
POL(app1_out(x₁, x₂)) = 2·x₁ + 2·x₂
POL(app2_in(x₁, x₂)) = 2·x₁ + 2·x₂
POL(app2_out(x₁)) = 2·x₁

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

PERM_IN(X) → U3¹(app1_in(X))
U4¹(app2_out(Z)) → PERM_IN(Z)

The TRS R consists of the following rules:

app2_in([], Y) → app2_out(Y)
app2_in(.(X), Y) → U2(app2_in(X, Y))
app1_in(.(Z)) → U1(app1_in(Z))
U2(app2_out(Z)) → app2_out(.(Z))
U1(app1_out(X, Y)) → app1_out(.(X), Y)

The set Q consists of the following terms:

app2_in(x0, x1)
app1_in(x0)
U2(x0)
U1(x0)

We have to consider all (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 2 less nodes.